# How do solve this pde problem?

EDIT: I know somehow, we end up with an equation relating the derivative of some coefficients to the rest of the stuff. I'm not sure where this equation, or even the constant that we use to get it, come from. Help with the derivation of this would be MUCH APPRECIATED I had these on an exam,and got zero credit for what I attempted, could someone suggest where to start? Please no full solutions.

Thank you.

You are given the Eigenvalues and eigenfunctions of the laplacian denoted $$\lambda_{i,j},\ \psi_{i,j}(\vec{r})$$respectively ( these are obtained from $$\nabla^2 u= \lambda u$$). Explain how you would solve the following with the initial condition $$u(\vec{r},t)=f_1(\vec{r})$$ applied to the PdE. $$\frac{\partial u}{\partial t}=\nabla^2 u$$

$$\frac{\partial u}{\partial t}+ b_{2,12} \psi_{2,12}= \nabla^2 u$$

Original Form of Question: You are given a complete set of eigenfunctions $$\phi _{n,k}$$ and eigenvalues$$\lambda _{n,k}$$ for the Laplacian on a domain R with homogeneous Dirichlet boundary conditions.The eigenfunctions are orthogonal with respect to a weight $$\sigma$$.

Explain how to solve the Heat Equation $$u_t=\nabla^2 u$$ on the domain R with the matching homogeneous boundary conditions with the intial condition $$u(r,0)= f_1(r)$$

.Explain how to solve $$u_t+b_{2,12}\phi_{2,12}(r,\theta )=\nabla^2 u$$ with $$u(0,r,\theta )=0$$ on the domain R with the matching homogeneous boundary conditions

• What are $b_{2,12}$ and $\psi_{2,12}$? Apr 30, 2016 at 0:10
• A constant, that may depend on time, and an eigenfunction with the index 2,12 Apr 30, 2016 at 0:12
• Ah right, okay. And the initial condition is $u(r,0) = f_1 (r)$? Apr 30, 2016 at 0:13
• Yes sir! Indeed Apr 30, 2016 at 0:13
• Recall that $\psi_{i,j}(r) e^{\lambda_{i,j}}$ is a solution to the first pde. To fit the initial conditions, you need to write $f_1$ as a linear combination of the $\psi_{i,j}$. For the second one, solve the homogenous part (with the $b_{2,12}$) first, then fix the boundary conditions. Apr 30, 2016 at 0:19

First the homogenous PDE, $\partial_t u = \Delta u$. As mentioned, $u_{i,j}(r,t) := \psi_{i,j}(r) e^{\lambda_{i,j} t}$ solves the heat equation, with initial condition at time zero $\psi_{i,j}(r)$. Since everything here is linear, a linear combination $\sum_{i,j} c_{i,j} u_{i,j}$ is also a solution. So the problem reduces to solving for the $c_{i,j}$ so that $f_1 = \sum c_{i,j} \psi_{i,j}$. Without any further knowledge of $f_1$, I am not sure what more can be said.
Now for the next one, try to find a particular solution to: $$\frac{\partial u}{\partial t}+ b_{2,12} \psi_{2,12}= \nabla^2 u$$ Hint: guess something of the form $\psi_{i,j}(r) \; T(t)$, which upon substitution into the PDE will become an ODE for $T(t)$. Then you can find a solution to the initial value problem by adding on the solution to the homogenous PDE.
• Just to check, we can say that $c_{i,j}= \frac{\int_{\mathbb{R}}(\sigma\ \psi_{i,j}f_1(x) )\mathbb{d}x}{\int_{\mathbb{R}}(\sigma\ \psi_{i,j}^2)\mathbb{d}x}$, correct?If so, why do we have to integrate, so that orthogonality cancels all $\psi_{i_1,j_1}\psi_{i_2,j_2}$ where $i_1 \neq i_2$,$j_1 \neq j_2$ Apr 30, 2016 at 2:21
• That's correct (although I don't know what $\sigma$ is). It seems you answer your question of why you have to integrate: in other words, substitute $f_1 = \sum c_{ij} \psi_{ij}$ into the right side of the equality and use orthogonality to check that you end up with the left side. Apr 30, 2016 at 2:27
• Okay, last thing, what about if the equation is $u_t + \gamma u = \nabla^2 u$?, The gamma is really messing with me. Sorry if i seem stupid... Apr 30, 2016 at 2:28
• It's actually easier in this case (compared to the 2,12), because it is still linear. Check that $\psi_{ij}e^{( \gamma- \lambda) t}$ are solutions; then same deal, find the $c$s to match initial condition. Apr 30, 2016 at 2:34